406 13. PROBLEMS LEADING TO ALGEBRAand whose length is 1 \ bu. This problem amounts to one linear equation in one
unknown. Dividing the area by the length yields 160 bu as the answer. The whole
difficulty of the problem lies in the complicated rules for dividing by a fraction.4.3. The Sun Zi Suan Jing. A large number of problems leading to algebra
are considered in the Sun Zi Suan Jing. Some of these are the kind of excess
and deficit problems already discussed. Others involve arithmetic and geometric
progressions and are solved by clever numerical reasoning. As an example of an
arithmetic progression, Problem 25 of Chapter 2 of the Sun Zi Suan Jing discusses
the distribution of 60 tangerines among five noblemen of different ranks in such a
way that each will receive three more than the one below him. Sun Zi says first
to give the lowest-ranking nobleman three, then six to the next-higher rank, and
so on, until the fifth person gets 15. That accounts for 3 + 6 + 9+12+15 = 45
tangerines and leaves fifteen more to be divided equally among the five. Thus the
numbers given out are 6, 9, 12, 15, and 18.4.4. Zhang Qiujian. To the fifth-century mathematician Zhang Qiujian (ca. 430-
490) we owe one of the most famous and long-lasting problems in the history of
algebra. It goes by the name of the Hundred Fowls Problem, and reads as follows:
Roosters cost 5 qian each, hens 3 qian each, and three baby chicks cost 1 qian.
If 100 fowls are bought for 100 qian, how many roosters, hens, and chicks were
bought? The answer is not unique, but Zhang gives all the physically possible
solutions: (4,18,78), (8,11,81), and (12,4,84). Probably this answer was obtained
by enumeration. Given that one is to buy at least one of each type of chicken, at
most 19 roosters can be bought. Zhang observed that the number of roosters must
increase in increments of 4, the number of hens must decrease in increments of 7,
and the number of baby chicks must increase in increments of 3. That is because
4-7 + 3 = 0 and 4·5-7·3 + 3- | = 0.
According to Mikami (1913, p. 41), three other "hardy perennials" of algebra
can be traced to Zhan Qiujian's treatise. One involves arithmetic progression. A
weaver produces 5 feet of fabric on the first day, and the output diminishes (by the
same amount) each day, until only 1 foot is produced on the thirtieth day. What
was the total production? The recipe for the answer is to add the amounts on the
first and last days, divide by 2, and multiply by the number of days.
The second is a rate problem of the type found in the Jiu Zhang Suanshu. A
horse thief rode 37 miles before his theft was discovered. The owner then pursued
him for 145 miles and narrowed the distance between them to 23 miles, but gave
up at that point and returned home. If he had continued the pursuit, how many
more miles would he have had to ride to catch the thief? Here we have the case
of one person traveling 145 miles in the same time required for the other to travel
131 miles, and the other person having a 23-mile head start. Following the formula
given in the Jiu Zhang Suanshu, Zhang Qiujian gives the answer as 145 ÷ 23 + 14.
Finally, we have another rate problem: If seven men construct 12^ bows in
nine days, how many days will be required for 17 men to construct 15 bows?
All these problems can be solved by reasoning about numbers without necessar-
ily writing down any equations. But they are definitely proto-algebra in that they
require thinking about performing operations on abstract, unspecified numbers.